Download Atomic Variational Calculations: Hydrogen to Boron

Simple Variational Calculations on Multi-electron Atoms and Ions
Frank Rioux
Department of Chemistry
College of St. Benedict | St. John's University
The purpose of this tutorial is to calculate the ground-state energies of simple multi-electron atoms and ions using
the variational method. In the interest of mathematical and computational simplicity the single parameter,
orthonormal hydrogenic wave functions shown below will be used. The method will be illustrated for boron, but
can be used for any atomic or ionic species with five or less electrons.
Using the following orthonormal trial wave functions, the various contributions to the total electronic energy of a
multi-electron atom are given below in terms of the variational parameter, α. For further detail see: "Atomic
Variational Calculations: Hydrogen to Boron," The Chemical Educator 1999, 4, 40-43. It should be pointed out
that in these calculations the exchange interaction is ignored. Including exchange generally improves the results by
about 1%.
α
Ψ 1s( α , r) :=
3
π
⋅ exp( −α ⋅ r)
Ψ 2s( α , r) :=
α
3
⎛ −α ⋅ r ⎞
⎟
⎝ 2 ⎠
⋅ ( 2 − α ⋅ r) ⋅ exp⎜
32⋅ π
α
Ψ 2p( α , r , θ ) :=
3
32⋅ π
⎛ −α ⋅ r ⎞ ⋅ cos( θ )
⎟
⎝ 2 ⎠
⋅ α ⋅ r⋅ exp⎜
The method will be illustrated with boron which has the electronic structure 1s 22s22p1.
Nuclear charge:
Seed value for variational parameter α:
Z := 5
Kinetic energy integrals:
T1s( α ) :=
α
2
2
T2s( α ) :=
α
2
T2p( α ) :=
8
α := Z
α
2
8
Electron-nucleus potential energy integrals:
VN1s( α ) := −Z⋅ α
Z
VN2s( α ) := − ⋅ α
4
Z
VN2p( α ) := − ⋅ α
4
Electron-electron potential energy integrals:
V1s1s( α ) :=
5
8
⋅α
V1s2s( α ) :=
17
81
⋅α
V1s2p( α ) :=
59
243
V2s2s( α ) :=
⋅α
77
512
⋅α
V2s2p( α ) :=
83
512
Enter coefficients for each contribution to the the total energy:
T1s
T2s
T2p
VN1s
VN2s
VN2p
V1s1s
V1s2s
V1s2p
V2s2s
V2s2p
a := 2
b := 2
c := 1
d := 2
e := 2
f := 1
g := 1
h := 4
i := 2
j := 1
k := 2
Variational energy equation:
E( α ) := a⋅ T1s( α ) + b ⋅ T2s( α ) + c⋅ T2p( α ) + d ⋅ VN1s( α ) + e⋅ VN2s( α ) + f⋅ VN2p( α ) ...
+ g ⋅ V1s1s( α ) + h ⋅ V1s2s( α ) + i⋅ V1s2p( α ) + j⋅ V2s2s( α ) + k ⋅ V2s2p( α )
Minimize energy with respect to the variational parameter, α.
α := Minimize( E , α )
α = 4.118
E( α ) = −23.320
⋅α
The experimental ground state energy is the negative of the sum of the successive ionization energies of
the atom or ion (see table of experimental data below):
SumIE := 0.305 + 0.926 + 1.395 + 9.527 + 12.500
E( α ) − Eexp
Compare theory and experiment:
Eexp
Eexp := −SumIE
Eexp = −24.653
= 5.405 %
Calculate orbital energies of a 1s, 2s and 2p electron by filling the place holders with the appropriate
coefficients (0, 1, 2, ...). Compare the calculated results with experimental values (see table below):
E1s( α ) := 1 ⋅ T1s( α ) + 0 ⋅ T2s( α ) + 0 ⋅ T2p( α ) + 1 ⋅ VN1s( α ) ...
+ 0 ⋅ VN2s( α ) + 0 ⋅ VN2p( α ) + 1 ⋅ V1s1s( α ) + 2 ⋅ V1s2s( α ) ...
+ 1 ⋅ V1s2p( α ) + 0 ⋅ V2s2s( α ) + 0 ⋅ V2s2p( α )
E1s( α ) = −6.809
Exp := −7.355
E2s( α ) := 0 ⋅ T1s( α ) + 1 ⋅ T2s( α ) + 0 ⋅ T2p( α ) + 0 ⋅ VN1s( α ) ...
+ 1 ⋅ VN2s( α ) + 0 ⋅ VN2p( α ) + 0 ⋅ V1s1s( α ) + 2 ⋅ V1s2s( α ) ...
+ 0 ⋅ V1s2p( α ) + 1 ⋅ V2s2s( α ) + 1 ⋅ V2s2p( α )
E2s( α ) = −0.012
Exp := −0.518
E2p( α ) := 0 ⋅ T1s( α ) + 0 ⋅ T2s( α ) + 1 ⋅ T2p( α ) + 0 ⋅ VN1s( α ) ...
+ 0 ⋅ VN2s( α ) + 1 ⋅ VN2p( α ) + 0 ⋅ V1s1s( α ) + 0 ⋅ V1s2s( α ) ...
+ 2 ⋅ V1s2p( α ) + 0 ⋅ V2s2s( α ) + 2 ⋅ V2s2p( α )
E2p( α ) = 0.307
Exp := −0.305
Successive Ionization Energies for the First Six Elements
⎛ Element
⎜
⎜ H
⎜ He
⎜ Li
⎜
⎜ Be
⎜ B
⎜
⎝ C
IE1
IE2
IE3
0.500
x
x
0.904 2.000
x
0.198 2.782 4.500
0.343 0.670 5.659
0.305 0.926 1.395
0.414 0.896 1.761
Oribtal Energies for the First Six Elements
⎞
⎟
x
x
x ⎟
x
x
x ⎟
⎟
x
x
x
⎟
8.000
x
x ⎟
9.527 12.500
x ⎟
⎟
2.370 14.482 18.000 ⎠
IE4
IE5
IE6
⎛ Element
⎜
⎜ H
⎜ He
⎜ Li
⎜
⎜ Be
⎜ B
⎜ C
⎝
1s
2s
−0.500
x
−0.904
x
−2.386 −0.198
−4.383 −0.343
−7.355 −0.518
−10.899 −0.655
⎞
⎟
x ⎟
x ⎟
x ⎟
⎟
x ⎟
−0.305 ⎟
⎟
−0.414 ⎠
2p
Radial Distribution Functions
2
2
2
2
r ⋅ Ψ 1s( α , r)
r ⋅ Ψ 2s( α , r)
2
r ⋅ Ψ 2p( α , r , 0)
2
3
0
0.5
1
1.5
2
r
2.5
3
3.5
Interpretation of results:
With this model for atomic structure we are able to compare theory with experiment in two ways. The calculated
ground-state energy is compared to the negative of the sum of the successive ionization energies. This
comparison shows that theory is in error by 5.4% - not bad for a one-parameter model for a five-electron atom.
However, the comparison of the calculated orbital energies with the negative of the orbital ionization energies is
not so favorable. It is clear that the one-parameter model used in this calculation does not do a very good job on
the valence electrons. For example, the 2s electrons are barely bound and the 2p electron is not bound at all.
The total energy (ground-state energy) comparison is more favorable because the model does a decent job on the
non-valence electrons where the vast majority of the energy resides. Chemistry, however, is dictated by the
behavior of the valence electrons, so the failure of the model to calculate good orbital energies, and therefore good
wave functions, for the valence electrons is a serious problem. This is a common problem in atomic and
molecular calculations; finding wave functions that effectively model the behavior of both the core and the
valence electrons.
Suggested additional problems: H, H -, Li, Li+, Be, C+, and Li atom excited state 1s22p1.